Introduction
A waiting period is more than a delay in payment. In a state-contingent contract, it often specifies a condition on the insured’s entire state history: the insured must remain in a qualifying state without interruption for a stated duration. A recovery, termination, or other disqualifying transition may restart the clock, even if the insured later returns to the qualifying state.
This distinction matters for valuation. A benefit deferred for one year from policy issue is not generally the same as a benefit payable after one uninterrupted year in disability. In the latter case, both the time of entry into disability and the probability of remaining disabled for the full waiting period affect whether and when payment occurs.
Turn the Clause Into a Payment Rule
Before calculating an expected present value, write down the event that triggers payment. A useful specification names the qualifying state, the length of the waiting period, what counts as an interruption, the benefit payment time, and the end of coverage. For a single benefit \(B\), payable at the completion of the waiting period, an eligible spell beginning at time \(s\) produces payment at \(s+w\) only if the insured remains in \(D\) throughout the interval from \(s\) to \(s+w\).
A state at just one instant cannot usually establish continuous occupancy over an interval. For example, in a continuous-time model, observing that the insured is disabled at both endpoints does not rule out recovery and re-entry between those endpoints. The calculation must reflect the path condition, not merely endpoint states.
Determine whether the waiting period starts at contract issue, at state entry, or at another specified event.
Identify which transitions end the qualifying spell. If any departure from the qualifying state resets the clock, a new spell must independently satisfy the full waiting period.
For a benefit paid at waiting-period completion, an entry at \(s\) can qualify by a coverage endpoint \(T\) only if \(s+w\leq T\).
Multiply the benefit by the probability of the required state history and discount to the actual payment time.
The examples below value a single trigger payment, rather than a stream of payments after eligibility. That separation makes the waiting-period rule explicit: the payment is made at the instant qualification is completed, if the contract provides for a lump sum at that instant.
Worked Example: Benefit After a Disability Spell Qualifies
A hypothetical contract begins with an insured in the active state \(A\). While active, disability begins with constant intensity \(\alpha=0.10\) per year. Once disabled, the insured exits disability through recovery or death, with total exit intensity \(q=0.20\) per year; either exit ends coverage. A benefit of \(\$6{,}000\) is paid after one uninterrupted year of disability. Coverage ends after \(T=3\) years. The force of interest is \(\delta=0.05\) per year. Find the expected present value at issue.
Method. Use the density of the first disability onset, the probability of remaining disabled through the waiting year, and discounting to the payment date. The model assumes constant transition intensities, no disability at issue, no recovery-and-reentry opportunity under the contract, and payment only if the waiting period finishes by year 3.
If disability begins at time \(s\), the onset density is \(\alpha e^{-\alpha s}\). The spell must start by \(T-w=3-1=2\), so \(0\leq s\leq2\). Given onset at \(s\), the probability of remaining disabled for one year is \(e^{-qw}=e^{-0.20}\). Payment occurs at \(s+w\), so its discount factor is \(e^{-\delta(s+w)}\). Therefore,
The calculation is checked by noting that onset must occur within two years and that the benefit is discounted for the full time \(s+1\), not merely for \(s\). The result is the expected present value of the qualifying lump sum, not the benefit amount times the probability of being disabled at year 3.
Why the Continuous-Time Formula Works
The preceding calculation follows from a general path-based argument. Suppose the insured starts active, can move from \(A\) to \(D\) with constant intensity \(\alpha\), and leaves \(D\) through an absorbing contract-ending state with constant total intensity \(q\). Let \(B\) be payable at the end of a continuous waiting period \(w\), let coverage end at \(T\), and let the force of interest be constant at \(\delta\). The model assumes there is at most one relevant disability spell: leaving \(D\) ends coverage, and the insured cannot re-enter.
For onset at time \(s\), the probability density of that onset is \(\alpha e^{-\alpha s}\). The conditional probability of no exit from \(D\) during the next \(w\) years is \(e^{-qw}\). The payment time is \(s+w\), with discount factor \(e^{-\delta(s+w)}\). An onset is eligible only when \(s+w\leq T\), or \(s\leq T-w\). Integrating the discounted payment over all possible eligible onset times gives the formula:
If \(T<w\), the waiting period cannot finish before coverage ends, so the value is zero. The factor \(e^{-qw}\) is essential: it is the probability that the insured survives in the qualifying state throughout the waiting period. The factor \(e^{-\delta w}\) accounts for discounting from onset to the later payment date. Together they appear as \(e^{-(q+\delta)w}\).
Worked Example: A Benefit for Someone Disabled at Issue
A hypothetical insured is already in qualifying state \(D\) at time 0. Recovery occurs with intensity \(0.25\) per year and death with intensity \(0.05\) per year. Either event prevents payment, and there are no other transitions. A \(\$10{,}000\) lump sum is payable after one uninterrupted year in \(D\). The force of interest is \(\delta=0.04\) per year. Find its value at time 0.
Method. Since the insured is already disabled at issue, there is no onset-time integral. The total exit intensity is \(q=0.25+0.05=0.30\) per year. The benefit is paid at time 1 only if no exit occurs during the year.
The probability of remaining in \(D\) for the full year is \(e^{-0.30(1)}=e^{-0.30}=0.7408182\). The discount factor to time 0 is \(e^{-0.04(1)}=e^{-0.04}=0.9607894\). Thus,
The calculation checks because the value is below \(\$10{,}000\): both the chance of remaining disabled and the time value of money reduce the expected present value. It would be incorrect to apply an onset intensity here, because disability is present at time 0.
Waiting Periods and Fixed Benefit Dates
Not every clause pays immediately when the waiting period is completed. A contract may instead specify a fixed assessment date and require that the insured have occupied the qualifying state throughout a particular interval before that date. In that case, the qualifying window and the payment date must be kept separate. The benefit is paid at the fixed date only for paths satisfying the full occupancy condition.
For a one-onset model with active-to-disabled intensity \(\alpha\), disability-exit intensity \(q\), and a fixed payment date \(T\), suppose payment requires disability throughout the final \(w\) years, from \(T-w\) through \(T\). An onset at \(s\leq T-w\) must be followed by no disability exit until \(T\). Its contribution to the probability of payment is the onset density \(\alpha e^{-\alpha s}\) multiplied by \(e^{-q(T-s)}\). The discounted benefit is then valued at time \(T\).
Worked Example: Qualification Over the Final Half-Year
A hypothetical contract starts with the insured active. Disability begins with intensity \(\alpha=0.20\) per year. Once disabled, the insured exits disability with intensity \(q=0.10\) per year, and an exit ends coverage. A \(\$10{,}000\) benefit is paid at the fixed time \(T=2\) years only if the insured has been continuously disabled throughout the final \(w=0.5\) year. The force of interest is \(\delta=0.03\) per year. Find the expected present value.
Method. An onset must occur by \(T-w=1.5\) years and the disability spell must then continue without exit through year 2. For an onset at \(s\), the density is \(0.20e^{-0.20s}\) and the conditional probability of no exit from \(s\) through 2 is \(e^{-0.10(2-s)}\). The benefit is discounted from the fixed payment time 2.
The payment-date discount factor is \(e^{-0.03(2)}=e^{-0.06}=0.9417645\). Therefore,
The probability is less than one because the insured must become disabled by year 1.5 and remain disabled until year 2. The payment is discounted for two years, not for \(s+0.5\), because this clause fixes the payment date at year 2.
Discrete Models and Contract Wording
A discrete-time model can apply the same logic when state changes are permitted only at specified observation times. Suppose the insured begins in \(D\), remains in \(D\) from one annual observation to the next with probability \(p\), and otherwise leaves \(D\) permanently. If the contract requires \(n\) uninterrupted annual periods in \(D\), the probability of qualification is \(p^n\), and a lump sum \(B\) paid at year \(n\) has value \(B p^n v^n\), where \(v=1/(1+i)\).
This calculation relies on the discrete model’s transition convention. If transitions can occur between annual observation dates, endpoint observations alone do not prove continuous occupancy. A continuous-time model or a more detailed contract-state specification is needed instead.
Worked Example: Two Annual Periods Without Interruption
A hypothetical annual model starts with an insured in qualifying state \(D\). At each year-end, a person in \(D\) remains there with probability \(0.80\); otherwise the person moves to a nonqualifying absorbing state and the contract ends. A benefit of \(\$5{,}000\) is payable at the end of year 2 only if the insured has remained in \(D\) through both annual periods. The effective annual interest rate is \(i=0.05\). Find the expected present value.
Method. The two annual continuations are required, so the probability of qualifying is \(0.80^2=0.64\). Payment occurs at year 2, so discount by \(v^2=(1.05)^{-2}\).
The result is checked by first finding the expected payment at year 2, \(\$5{,}000(0.64)=\$3{,}200\), then discounting it two years. This discrete calculation treats the two annual transitions as the only possible changes in state; it does not assert that observing disability at the two year-ends establishes continuous disability between them.
Common Mistakes and Exam Checks
- Starting the clock at the wrong event. A one-year waiting period after disability onset is not a one-year deferral from policy issue. Use the clause’s specified clock start.
- Ignoring exits during the waiting period. In a continuous-time model, the insured must remain in the qualifying state for the whole interval. Include the probability of no disqualifying transition during that time.
- Discounting to the wrong date. If onset is at \(s\) and payment is at \(s+w\), discount to \(s+w\). If the clause specifies a fixed payment date \(T\), discount to \(T\).
- Using only the state at the endpoint. Being in \(D\) at time \(s+w\) does not establish uninterrupted occupancy over the preceding \(w\) years unless the model and contract rule make that conclusion valid.
- Assuming interrupted time accumulates. A continuous-occupancy clause resets after a disqualifying exit. Accumulation or credit for earlier spells must be explicitly provided by the contract.
For the first worked example, a common incorrect calculation is to omit the probability of remaining disabled during the one-year wait. That calculation is
It discounts the one-year delay but treats the qualifying spell as if exit during that year were impossible. The correct calculation also multiplies by \(e^{-0.20(1)}\), giving \(\$986.16e^{-0.20}=\$807.40\). Always separate the survival condition from the discount factor: they are different parts of the valuation.
Check Your Understanding
For each problem, interpret the waiting rule before calculating. Assume the stated transition models and contract terms apply exactly as given.
- An insured is in \(D\) at issue. The total exit intensity from \(D\) is \(0.25\) per year. A \(\$8{,}000\) benefit is paid after two uninterrupted years in \(D\), and \(\delta=0.03\) per year. Find the expected present value. (Answer: \(\$4{,}569.67\))
- An annual model starts in \(D\). At each year-end, the probability of remaining in \(D\) is \(0.80\); otherwise the insured exits permanently. A \(\$10{,}000\) benefit is paid after two uninterrupted annual periods in \(D\). The effective annual interest rate is \(25\%\). Find the expected present value. (Answer: \(\$4{,}096.00\))
- A contract starts active. Disability onset has intensity \(0.10\) per year; after onset, the total exit intensity is \(0.20\) per year and an exit ends coverage. A \(\$5{,}000\) benefit is paid after \(0.5\) uninterrupted year in disability, coverage ends at \(T=2\), and \(\delta=0.05\) per year. Find the expected present value. (Answer: \(\$592.70\))
- An insured starts in \(D\). In an annual model, the probability of remaining in \(D\) at each year-end is \(0.75\), with any exit permanent. A \(\$4{,}000\) benefit is paid after two uninterrupted years in \(D\). The effective annual interest rate is \(10\%\). Find the expected present value. (Answer: \(\$1{,}859.50\))
- A contract starts active. Disability onset has intensity \(0.20\) per year, and disability exit has intensity \(0.10\) per year and ends coverage. A \(\$10{,}000\) benefit is payable at \(T=2\) only if disability has been continuous throughout the final \(0.5\) year. With \(\delta=0.03\), find the expected present value. (Answer: \(\$2{,}148.03\))